Question

how do i find the sum of the infinite series ((-2)^n)/(e^(n+1))?

Answer 1

\[\sum_{n =1}^{\infty} \frac{ (-2)^{n} }{ e^{n+1} }\]

Answer 2

\[\sum_{n =1}^{\infty} \frac{ (-2)^{n} }{ e^{n+1} } = \dfrac{1}{e}\sum_{n =1}^{\infty} \frac{ (-2)^n}{ e^n }=\dfrac{1}{e}\sum_{n =1}^{\infty} \left(\dfrac{-2}{e}\right)^{n} \]Yes?

Answer 3

How would I go from there?

Answer 4

Familiar with formula for sum of an infinite geometric series?

Answer 5

e > 2 so you have a geometric series with |r| < 1

Answer 6

Oh... I see. Thanks!

Answer 7

Be careful; the formula is \[\sum_{n=0}^\infty r^n = \dfrac{1}{1-r}\]
Here, you have bottom limit of summation \(n=1\) instead of \(n=0\).

Answer 8

And welcome.